+0

# Help? Complex numbers

0
43
2

Help? Complex numbers

Guest Oct 3, 2017

#2
+90639
+2

I think you have misunderstood the question guest ://

j is not a complex number.

x^2-3x+2j=0

Solutions for x

$$x=\frac{3\pm\sqrt{9-8j}}{2}$$

The roots will be complex when  the discriminant is negative.

That is

$$9-8j<0\\ 9<8j\\ 9/8 9/8\\ j>1.125$$

So the solutions (for x) of  x^2-3x+2j=0    will be complex if   j>1.125

Melody  Oct 3, 2017
Sort:

#1
+1

I replaced your "j" with "i", which is a more commonly used "complex number"

Solve for x:
2 i - 3 x + x^2 = 0

Subtract 2 i from both sides:
x^2 - 3 x = -2 i

x^2 - 3 x + 9/4 = 9/4 - 2 i

Write the left hand side as a square:
(x - 3/2)^2 = 9/4 - 2 i

Take the square root of both sides:
x - 3/2 = sqrt(9/4 - 2 i) or x - 3/2 = -sqrt(9/4 - 2 i)

x = sqrt(9/4 - 2 i) + 3/2 or x - 3/2 = -sqrt(9/4 - 2 i)

x = 3/2 + 1/2 sqrt(9 - 8 i)      or      x = 3/2 - sqrt(9/4 - 2 i)

Guest Oct 3, 2017
#2
+90639
+2

I think you have misunderstood the question guest ://

j is not a complex number.

x^2-3x+2j=0

Solutions for x

$$x=\frac{3\pm\sqrt{9-8j}}{2}$$

The roots will be complex when  the discriminant is negative.

That is

$$9-8j<0\\ 9<8j\\ 9/8 9/8\\ j>1.125$$

So the solutions (for x) of  x^2-3x+2j=0    will be complex if   j>1.125

Melody  Oct 3, 2017

### 20 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details